POJ 2075 Tangled in Cables

链接：http://poj.org/problem?id=2075

题意：给N个城市，用电缆将所有城市连接起来。

思路：最小生成树模板题,kruskal算法的应用。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<ctype.h>
#include<algorithm>
#include<string>
#define PI acos(-1.0)
#define maxn 600
#define INF 1<<25
#define MAX 0x7fffffff
typedef long long ll;
using namespace std;
string c[maxn];
int vv[maxn];
struct Road
{
int v,w;
double l;
} road[maxn*maxn];
bool cmp(Road a,Road b)
{
return a.l<b.l;
}
int findset(int x)
{
if(x==vv[x])
return x;
return vv[x]=findset(vv[x]);
}
int unionset(int x,int y)
{
int xx=findset(x);
int yy=findset(y);
if(xx==yy)
return 0;
else if(yy>xx)
{
vv[yy]=xx;
}
else vv[xx]=yy;
return 1;
}
int main()
{
double len,k,ans=0,tt=0;
string x,y;
int cc,rr,a,b;
memset(vv,0,sizeof(vv));
scanf("%lf",&len);
scanf("%d",&cc);
for(int i=1; i<=cc; i++)
{
vv[i]=i;
cin>>c[i];
}
scanf("%d",&rr);
for(int ii=0; ii<rr; ii++)
{
cin>>x>>y>>k;
for(int i=1; i<=cc; i++)
{
if(c[i]==x)
a=i;
if(c[i]==y)
b=i;
}
road[ii].v=a;
road[ii].w=b;
road[ii].l=k;
}
sort(road,road+rr,cmp);
for(int i=0; i<rr; i++)
{
if(unionset(road[i].v,road[i].w))
{
ans+=road[i].l;
tt++;
}
if(tt==cc-1)
break;
}
if(ans<=len)
printf("Need %.1lf miles of cable\n",ans);
else printf("Not enough cable\n");
}